Ggtcf

Here I was, innocently trying to solve this daunting-looking integral

$$int_0^pi e^v cos theta cos t cosh(v sin theta sin t) dt $$

when the inner beauty behind this beast slowly started to disclose itself.

Of course, the first thing I did, was checking if WolframAlpha can help, futilely.

Next on my lookup table was Gradshteyn and Ryzhik. Yet again, the god of integrals had no mercy upon me and I was left to my own device.

To get a first impression I plotted the function, that needs to be integrated, that is for $f_theta(t) = e^cos theta cos t cosh(sin theta sin t)$, we get the following plots:

Okay, nothing too special about that. So I proceeded by trying to numerically evaluate the integral itself. And then, something strange happened...

Turns out the integral is invariant under $theta$!

Even better, we have

$$int_0^pi e^v cos theta cos t cosh(v sin theta sin t) dt = int_0^pi e^v cos t dt = pi I_0(v)quad forall theta in [-pi,pi]; ,$$

where for the first equality I just set $theta = 0$ and the second equality is a known identity of the modified Bessel function of the first kind. Now, I only stumbled upon this identity numerically, and I was wondering if someone can share some analytical wisdom regarding this. Put into a question:

Does someone know, why this identity holds?

Bonus

I now face the same integral but with an additional linear term, that is

$$int_0^pi tf_theta(t)dt ; ,$$
with $f_theta$ as defined above. I am hoping that the techniques that illuminate the identity above will also shed some light at this new integral, which by the way is not constant in $theta$ anymore.

For the main problem, a bit of algebra:
beginalign*e^vcosthetacos tcosh(vsinthetasin t) &= e^vcosthetacos tleft(e^vsinthetasin t+e^-vsinthetasin tright)\
&= frac12left(e^vcosthetacos t+vsinthetasin t+e^vcosthetacos t-vsinthetasin tright)\
&= frac12left(e^vcos(theta-t)+e^vcos(theta+t)right)endalign*
Now we integrate that:
beginalign*int_0^pie^vcosthetacos tcosh(vsinthetasin t),dt &= frac12int_0^pie^vcos(theta-t)+e^vcos(theta+t),dt\
&=frac12left(int_0^pie^vcos(theta+t),dt+int_-pi^0e^vcos(theta+s),dsright)\
&=frac12int_-pi^pie^vcos(theta+t),dt = frac12int_-pi-theta^pi-thetae^vcos s,dsendalign*
Flipping $theta-t$ to $theta+s$ gives us an integral over the other half of the period - and it's the same function, so we just write it as one integral. Then, in that final integral of $e^vcos s$ over one full period, it doesn't matter where that period is; from $-pi$ to $pi$ is the same as from $-pi-theta$ to $pi-theta$.

That leaves us with the Bessel function identity, that the average value of $e^vcos t$ over a full period is $I_0(cos v)$. For this, since Bessel functions are defined by a differential equation, we differentiate (under the integral sign):
beginalign*I(v) &= frac12piint_0^2pie^vcostheta,dtheta\
I'(v) &= frac12piint_0^2picosthetacdot e^vcostheta,dtheta\
I''(v) &= frac12piint_0^2picos^2thetacdot e^vcostheta,dtheta\
I'(v) &= frac12pileft[sinthetacdot e^vcosthetaright]_theta=0^theta=2pi +frac12piint_0^2pisinthetacdot vsinthetacdot e^vcostheta,dtheta\
I'(v) &= fracv2piint_0^2pisin^2thetacdot e^vcostheta,dthetaendalign*
The first three lines are $I$ and its derivatives, calculated the obvious way. Then, in the next two, we apply integration by parts to transform the $I'$ integral into a form that works better with the others. Then, from $cos^2+sin^2=1$, we get $vI''(v)+I'(v)-vI(v)=0$, the modified Bessel equation of order zero. Together with the initial condition $I(0)=1$ (since the average value of $1$ is $1$) and $I'(0)=0$, this gives that $I(v)=I_0(v)$. Done.

A brief note on the bonus question: we can apply the same identities, but we run into trouble when we try to fold over and transform the $e^vcos(theta-t)$ term into an integral over $[-pi,0]$. The way the $t$ factor transforms, we end up with
$$frac12int_-pi^pi|t|cdot e^vcos(theta+t),dt$$
Multiplying by a triangle wave isn't going to come out cleanly. I might look at Fourier series next, but not in this answer.

Note that
$$beginalign*
I_theta &= frac14int_0^2pie^vcosthetacos tleft(e^vsinthetasin t+e^-vsinthetasin tright) mathrm dt\&=frac14int_0^2pie^vcos(t-theta) mathrm dt+frac14int_0^2pie^vcos(t+theta) mathrm dt.
endalign*$$ Because $tmapsto e^vcos t$ is $2pi$-periodic, we have
$$
int_0^2pie^vcos(t-theta) mathrm dt=int_0^2pie^vcos(t+theta) mathrm dt=int_0^2pie^vcos t mathrm dt
$$ so it follows that
$$
I_theta = frac 12int_0^2pie^vcos t mathrm dt =int_0^pi e^vcos t mathrm dt.
$$